Question

The energy required to melt 1 g ice is 33J. The number of quanta of radiation of frequency $\text{4}\text{.67}\times {\text{10}}^{\text{13}}{\sec }^{-1}$ that must be absorbed to melt 10 g ice is:

• A. $1.067\times {10}^{22}$
• B. $3.205\times {10}^{23}$
• C. $9.076\times {10}^2$
• D. None of the above

Answer 1

The correct option is A $1.067\times {10}^{22}$

Energy required to melt $1g$ ice $=33J$.

Energy required to melt $10g$ ice $=33\times 10=330J$.

Radiation frequency $\left(\nu \right)=4.67\times {10}^{13}{s}^{-1}$

As we know, the energy of radiation is given by:

$E=h\nu$

Here, $h=$ Planck's constant $=6.62\times {10}^{-34}J.s$

$\therefore E=6.62\times {10}^{-34}\times 4.67\times {10}^{13}$

$\Rightarrow E=3.09\times {10}^{-20}J$

$\therefore$ No. of quanta required $=\frac{330}{3.09\times {10}^{-20}}=1.067\times {10}^{22}$

Hence, the correct option is $\left(A\right)$.